Determine the speed of a falling object 7.8 sec after it dropped.

Physics

1 answer:

butalik [34]3 years ago

7 0

Answer:

76.4m/s

Explanation:

Given parameter:

Time taken = 7.8sec

Unknown:

Speed after it dropped = ?

Solution:

To solve this problem, we use one of the kinematics equation:

V = U + gt

V is the final speed

U is the initial speed = 0m/s

g is the acceleration due to gravity

t is the time taken

V = 0 + 9.8 x 7.8 = 76.4m/s

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If the absolute temperature of a gas is 600 K, the temperature in degrees Celsius is: A. 705Â°C. B. 873Â°C. C. 273Â°C. D. 327Â°C

zlopas [31]

Answer:

D). $327 ^0 C$

Explanation:

As we know that temperature scale is linear so we will have

$\frac{^0C - 0}{100 - 0} = \frac{K - 273}{373 - 273}$

now we have

$\frac{^0 C - 0}{100} = \frac{K - 273}{100}$

so the relation between two scales is given as

$^0 C = K - 273$

now we know that in kelvin scale the absolute temperature is 600 K

so now we have

$T = 600 - 273 = 327 ^0 C$

so correct answer is

D). $327 ^0 C$

4 0

3 years ago

An object is moving to the west at a constant speed. three forces are exerted on the object. one force is 10 n directed due nort

SVETLANKA909090 [29]

If we want the object to continue to move at constant speed, it means that the resultant of the forces acting on the object must be zero. So far, we have:
- force F1 with direction north, of 10 N
- force F2 with direction west, of 10 N
The third force must balance them, in order to have a net force of zero on the object.

The resultant of the two forces F1 and F2 is
$F_{12} = \sqrt{F_1^2+F_2^2}= \sqrt{(10 N)^2+(10 N)^2}= \sqrt{200}=14.1 N$
with direction at $45^{\circ}$ north-west. This means that F3 must be equal and opposite to this force: so, F3 must have magnitude 14.1 N and its direction should be $45^{\circ}$ south-east.